Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises: 23

Answer

$$\int\tan^2 xdx=\tan x-x+C$$

Work Step by Step

$$A=\int\tan^2 xdx$$ Remember that $\sec^2 x=1+\tan^2 x$. So, $\tan^2 x=\sec^2 x-1$ $$A=\int(\sec^2x-1)dx$$ Also, $\int\sec^2 x=\tan x+C$. Therefore, $$A=\tan x-x+C$$
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